Given: The systems of equation below
[tex]\begin{gathered} 4x+8y=68 \\ 12x+3y=39 \end{gathered}[/tex]To Determine: The solution of the system of equations using elimination method
Eliminate x
To eliminate x, multiply the first equation by 3 and the second equation by 1
[tex]\begin{gathered} 3\times(4x+8y=68)=12x+24y=204 \\ 1\times(12x+3y=39)=12x+3y=39 \end{gathered}[/tex]Combine two equations by subtracting the second equation from the first
[tex]\begin{gathered} 12x-12x+24y-3y=204-39 \\ 21y=165 \\ \frac{21y}{21}=\frac{165}{21} \\ y=\frac{165}{21}=\frac{55}{7} \end{gathered}[/tex]Substitute y in the first equation to get x
[tex]\begin{gathered} 4x+8(\frac{55}{7})=68 \\ 4x+\frac{440}{7}=68 \\ 4x=68-\frac{440}{7} \\ 4x=\frac{476-440}{7} \\ 4x=\frac{36}{7} \\ x=\frac{36}{7}\times\frac{1}{4} \\ x=\frac{9}{7} \end{gathered}[/tex]Hence, the solution is
[tex](\frac{9}{7},\frac{55}{7})[/tex](9/7 , 55/7)
